Staff: Mentor

Couldn't I just inset 0 for x and end up getting [itex]\sqrt{1}[/itex]?

The book tells me to rewrite this problem by the conjugate.

My question is why? Thanks

As you have written it,
[tex]\lim_{x \to \infty}\sqrt{x^2 + 1} - 1 = \infty[/tex]
You seem to be a little confused in this problem. Why would you put in zero for x when the limit is as x goes to infinity?